d Inverse Trigonometric Functions i E 1 Lecture a l l u d b - - PowerPoint PPT Presentation

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Section 3.9 Inverse Trigonometric Functions 1 Lecture

• Dr. Abdulla Eid

College of Science

MATHS 101: Calculus I

• Dr. Abdulla Eid (University of Bahrain)

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Inverse Trigonometric functions

1 The derivative of the basic inverse trigonometric functions. 2 The derivative of functions that involve inverse trigonometric

functions.

3 Identities of inverse trigonometric functions using differential calculus.

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f (x) = sin x has an inverse if x ∈ [ −π

2 , π 2 ]. The inverse of the sine

function is denoted by f −1(x) = sin−1 x (= arcsin x) with the following properties:

1 The graph of y = sin−1 x is: 2 The domain of sin−1 x is [−1, 1] 3 The range of sin−1 x is [ −π

2 , π 2 ]

4 sin

• sin−1 x

= x

5 sin−1 (sin x) = x

Exercise 1

Write a similar definition for the tan−1 x and sec−1 x and explore their properties.

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Derivative of sin−1 x

Example 2

Find

d dx

• sin−1 x
• .

Solution: y = sin−1 x, find y ′ sin y = sin

• sin−1 x
• sin y = x

cos y · y ′ = 1 y ′ = 1 cos y = sec y sin y = x = x 1 = opp hypo (adj)2 + (opp)2 = (hypo)2 (adj)2 + x2 = 1 (adj)2 = 1 − x2 adj =

• 1 − x2

sec y = hypo adj = 1 √ 1 − x2 y ′ = d dx

• sin−1 x

= 1 √ 1 − x2

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Exercise 3

Derive a formula for

d dx

• cos−1 x
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Derivative of tan−1 x

Example 4

Find

d dx

• tan−1 x
• .

Solution: y = tan−1 x, find y ′ tan y = tan

• tan−1 x
• tan y = x

sec2 y · y ′ = 1 y ′ = 1 sec2 y = cos2 y tan y = x = x 1 = opp adj (adj)2 + (opp)2 = (hypo)2 1 + x2 = (hypo)2 (hypo)2 = 1 + x2 hyp =

• 1 + x2

cos y = adj hypo = 1 √ 1 + x2 y ′ = d dx

• tan−1 x

= 1 1 + x2

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Exercise 5

Derive a formula for

d dx

• cot−1 x
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Derivative of sec−1 x

Example 6

Find

d dx

• sec−1 x
• .

Solution: y = sec−1 x, find y ′ tan y = sec

• sec−1 x
• sec y = x

sec y tan y · y ′ = 1 y ′ = 1 sec y tan y = cos y cot y sec y = x = x 1 = hypo adj (adj)2 + (opp)2 = (hypo)2 1 + (opp)2 = x2 (opp)2 = x2 − 1

• pp =
• x2 − 1

cot y = adj

• pp =

1 √ x2 − 1 y ′ = d dx

• sec−1 x

= 1 |x| √ x2 − 1

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Exercise 7

Derive a formula for

d dx

• csc−1 x
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Summary

(Derivative of the Inverse Trigonometic functions) (sin−1 x)′ = 1 √ 1 − x2 (cos−1 x)′ = −1 √ 1 − x2 (tan−1 x)′ = 1 1 + x2 (cot−1 x)′ = −1 1 + x2 (sec−1 x)′ = 1 |x| √ x2 − 1 (csc−1 x)′ = −1 |x| √ x2 − 1

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2 - Derivative of functions that involve the inverse trigonometric functions

Example 8

Find y ′ if y = tan−1(√x). Solution: We have y = tan−1(√x) y ′ = 1 1 + (√x)2 ·

• 1

2√x

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Exercise 9

Find y ′ if y = √ tan−1 x. Solution: We have y = √ tan−1 x y ′ = 1 2 √ tan−1 x ·

• 1

1 + x2

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Example 10

Find y ′ if y = cos−1 (ex). Solution: We have y = cos−1 (ex) y ′ = −1

• 1 − (ex)2 · (ex)
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Exercise 11

Find y ′ if y = arcsin (3 − 2x). Solution: We have y = arcsin (3 − 2x) y ′ = 1

• 1 − (3 − 2x) 2 · (−2)
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Example 12

Find y ′ if y = sin−1 √ sin x

• .

Solution: We have y = sin−1 √ sin x

• y ′ =

1

• 1 −

√ sin x

• 2

· √ sin x ′ = 1

• 1 − (

√ sin x)2 · 1 2 √ sin x · cos x = 1 √ 1 − sin x · 1 2 √ sin x · cos x

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Exercise 13

Find y ′ if y = x sin−1 x + √ 1 − x2 and simplify your answer. Solution: y ′ = (1) sin−1 x + x · 1 √ 1 − x2 + 1 2 √ 1 − x2 · (−2x) y ′ = sin−1 x + x √ 1 − x2 − x √ 1 − x2 = sin−1 x

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Identities of inverse trigonometric functions using differential calculus.

Example 14

Find y ′ if y = sin−1 + cos−1 x. Solution: y ′ = 1 √ 1 − x2 + −1 √ 1 − x2 y ′ = 0 The derivative is always zero, which means the function y is a constant. y = sin−1 + cos−1 x = C → sin−1(0) + cos−1(0) = C → π 2 = C Hence the identity is sin−1 + cos−1 x = π 2

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Exercise 15

Show using calculus that

1 cot−1 x = π

2 − tan−1 x

2 csc−1 x = π

2 − sec−1 x

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Example 16

Find y ′ if y = tan−1 x + tan−1 1

x

• .

Solution: y ′ = 1 1 + x2 + 1 1 + 1

x

2 · −1 x2 → 1 1 + x2 + x2 1 + x2 · −1 x2 y ′ = 1 1 + x2 − 1 1 + x2 → y ′ = 0 The derivative is always zero, which means the function y is a constant. y = tan−1 x + tan−1 1 x

• = C → tan−1(1) + tan−1

1 1

• = C → π

2 = C Hence the identity is tan−1 x + tan−1 1 x

• = π

2

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Exercise 17

Show that sec−1(−x) + sec−1(x) = π.

Exercise 18

(At home) Show that tan−1 x + tan−1 a = tan−1 x + a 1 − ax

• and use it to compute tan−1 1 + tan−1 2 + tan−1 3.

Exercise 19

(Challenging problem) What is wrong in d dx

• sin−1 x + sec−1 x

= 1 √ 1 − x2 + 1 |x| √ x2 − 1

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